A heat engine is involved with exchange of heat

(3) Efficiency, $\eta=\frac{\text { Work done }}{\text { Heat absorbed }}=\frac{W}{\Sigma Q}$

$=\frac{Q_{1}+Q_{2}+Q_{3}+Q_{4}}{Q_{1}+Q_{3}}=0.5$

Here, $Q_{1}=1915 \mathrm{~J}, Q_{2}=-40 \mathrm{~J}$ and $Q_{3}=125 \mathrm{~J}$

$\therefore \frac{1915-40+125+Q_{4}}{1915+125}=0.5$

$\Rightarrow 1915-40+125+Q_{4}=1020$

$\Rightarrow Q_{4}=1020-2000$

$\Rightarrow Q_{4}=-Q=-980 \mathrm{~J} \Rightarrow Q=980 \mathrm{~J}$